2 years ago

I need help with 1 and 2 please :)

Mathematics

1 answer:

marin [14]2 years ago

8 0

#1 is the third one down (x+3)^2 + (y-2)^2 = 9

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2 years ago

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IceJOKER [234]

$\bf 343^{\frac{2}{3}}+36^{\frac{1}{2}}-256^{\frac{3}{4}}\qquad \begin{cases}
343=7\cdot 7\cdot 7\\
\qquad 7^3\\
36=6\cdot 6\\
\qquad 6^2\\
256=4\cdot 4\cdot 4\cdot 4\\
\qquad 4^4
\end{cases}\\\\\\ (7^3)^{\frac{2}{3}}+(6^2)^{\frac{1}{2}}-(4^4)^{\frac{3}{4}}
\\\\\\
\sqrt[3]{(7^3)^2}+\sqrt[2]{(6^2)^1}-\sqrt[4]{(4^4)^3}\implies \sqrt[3]{(7^2)^3}+\sqrt[2]{(6^1)^2}-\sqrt[4]{(4^3)^4}
\\\\\\
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to see what you can take out of the radical, you can always do a quick "prime factoring" of the values, that way you can break it in factors to see who is what.

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3 years ago

(- 1, 2) and (9, - 3)

slava [35]

9/5

the distance is 9 over, and 5 up.

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3 years ago

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